Researchers thought this was a bug (Borwein integrals)

Q: Why does multiplying more sinc factors not immediately change the integral’s value away from $\pi$?

Each added factor is $\mathrm{sinc}(x/k)=\frac{\sin(\pi x/k)}{\pi x/k}$, which is less than 1 for most $x\neq 0$. Naively that should shrink the signed area. The surprise is that, under Fourier-transform duality, the product of sinc terms corresponds to repeated convolution/averaging of a rectangular (“rect”) function. As long as the effective averaging windows fit entirely within the rect plateau, the global quantity (the integral) remains pinned—so the area stays exactly $\pi$ for many steps before the plateau is finally eroded.

Q: What is the moving-average analogy, and how does it predict when the value drops?

Begin with $\mathrm{rect}(x)$, equal to 1 on $[-1/2,1/2]$ and 0 outside. Define $f_2$ as the moving average of $f_1$ over a window of width $1/3$; at $x=0$, the output stays exactly 1 as long as the window lies fully inside the plateau where the function equals 1. Each new step averages over a smaller odd reciprocal window ($1/5,1/7,\dots$), shrinking the plateau by that window width. The value at $x=0$ first drops when the remaining plateau becomes thinner than the next averaging window—equivalently when the cumulative sum of odd reciprocals exceeds the plateau length.

TL;DR

The Borwein sinc-product integrals equal $π$ exactly for many steps, then drop slightly below $π$ only after a specific threshold index (15 in the basic sequence, 113 when $2 cos (x)$ is included).

Briefing Cornell Notes

Briefing

A family of integrals built from the “engineer sinc” function— $sinc (x) = \frac{s i n ( π x )}{π x}$ —keeps landing exactly on $π$ for a surprisingly long run, even after repeatedly multiplying in more stretched sinc factors. The catch is that the equality eventually fails, but only after a threshold: with the classic Borwein-style sequence it holds until a specific odd stretch factor (15), and with an added $2 cos (x)$ term it holds until much later (113). The size of the eventual drop is so tiny that it can look like a numerical glitch—until the exact arithmetic reveals it’s a real, structured phenomenon.

The integrals start with the signed area under $\frac{s i n x}{x}$ from $- \infty$ to $\infty$ , which equals $π$ . Then the sequence multiplies in additional sinc terms, each time stretching the argument by an odd number: first $sinc (x /3)$ , then $sinc (x /5)$ , and so on. Intuition suggests the area should shrink because each factor is less than 1 away from $x = 0$ , yet the area stubbornly stays $π$ for many steps. When it finally breaks, the exact value is a fraction of $π$ with an astronomically large numerator and denominator, ruling out floating-point error.

This behavior traces back to a paper by Jonathan Borwein and David Borwein, who noted that a colleague using a computer algebra system dismissed the pattern as a bug. It wasn’t. The integrals remain exactly $π$ as long as a related “budget” stays under control; once the budget is exceeded, the value slips below $π$ by a barely perceptible amount.

To make that budget concrete, the video builds an analogy using moving averages of a step function. Start with a rectangle (a “rect” function) that equals 1 on $[- 1/2, 1/2]$ and 0 outside. Each new function is formed by averaging the previous one over a shrinking window whose width is an odd reciprocal ( $1/3, 1/5, 1/7, \dots$ ). At $x = 0$ , the output stays exactly 1 as long as the averaging window fits entirely inside the plateau where the function equals 1. The plateau shrinks by the window width each round; the first time the window becomes wider than the remaining plateau, the value at $x = 0$ drops slightly below 1.

The “break index” matches the integral’s break index because the plateau survives until the sum of the odd reciprocals exceeds the initial plateau length. For the $1$ -length plateau, the threshold occurs at $1/3 + 1/5 + \dots + 1/15 > 1$ , while for a longer plateau (length 2, corresponding to the extra $2 cos (x)$ factor) the threshold is delayed until $113$ . The same mechanism—plateau erosion under repeated averaging—explains why the integral stays locked to $π$ for so long, then finally slips.

The final bridge between the two worlds comes from Fourier transforms and the convolution theorem. The sinc/rect relationship under Fourier transform turns “integral of sinc products” into “repeated averaging” in a dual domain. In that language, multiplying sinc factors corresponds to convolving (i.e., averaging) rect-shaped objects, so the integral’s stability becomes a statement about how long a rectangular plateau persists under successive windowing. The result isn’t just a curiosity: it’s a clean example of how shifting perspectives with Fourier analysis can turn a hard global quantity (an infinite signed area) into something governed by simple geometric constraints.

Cornell Notes

The Borwein-style integrals built from $sin x / x$ stay exactly equal to $π$ for many steps, even as more stretched sinc factors are multiplied in. The equality fails only when a threshold is crossed: for the basic sequence it breaks at 15, and when an extra $2 cos (x)$ factor is included it holds until 113. A parallel moving-average model explains the timing: start with a step (“rect”) function, repeatedly average it over windows of widths $1/3, 1/5, 1/7, \dots$ , and track the value at $x = 0$ . The value stays at the plateau height until the cumulative window widths eat through the plateau; the break happens when the sum of the chosen reciprocals exceeds the plateau length. Fourier transforms and the convolution theorem connect this averaging picture to the sinc integral products.

Why does multiplying more sinc factors not immediately change the integral’s value away from $π$ ?

Each added factor is

sinc (x / k) = \frac{s i n ( π x / k )}{π x / k}

, which is less than 1 for most

x \neq = 0

. Naively that should shrink the signed area. The surprise is that, under Fourier-transform duality, the product of sinc terms corresponds to repeated convolution/averaging of a rectangular (“rect”) function. As long as the effective averaging windows fit entirely within the rect plateau, the global quantity (the integral) remains pinned—so the area stays exactly

π

for many steps before the plateau is finally eroded.

What is the moving-average analogy, and how does it predict when the value drops?

Begin with

rect (x)

, equal to 1 on

[- 1/2, 1/2]

and 0 outside. Define

f_{2}

as the moving average of

f_{1}

over a window of width

1/3

; at

x = 0

, the output stays exactly 1 as long as the window lies fully inside the plateau where the function equals 1. Each new step averages over a smaller odd reciprocal window (

1/5, 1/7, \dots

), shrinking the plateau by that window width. The value at

x = 0

first drops when the remaining plateau becomes thinner than the next averaging window—equivalently when the cumulative sum of odd reciprocals exceeds the plateau length.

How does the “break at 15” connect to a sum of reciprocals of odd numbers?

In the rect/averaging model, the initial plateau length is 1. Each averaging step with window width

1/ (2 n + 1)

reduces the plateau by that amount (in the sense that the window must fit inside the plateau to keep the average at 1). The pattern holds until

1/3 + 1/5 + 1/7 + \dots

reaches a point where the total eaten width exceeds 1. In the highlighted Borwein sequence, that first happens at the stage corresponding to 15, so the integral’s exact

π

equality persists up to that index and then slips slightly below.

Why does adding a $2 cos (x)$ factor delay the breakdown until 113?

The extra factor changes the dual-domain starting shape: the rect plateau effectively becomes longer (from length 1 to length 2). In the moving-average analogy, a longer plateau means the successive windows must “eat” through more total width before the average at the center can drop below the plateau height. That pushes the threshold to a later odd reciprocal stage; the cumulative odd-reciprocal sum doesn’t exceed 2 until the index corresponding to 113, matching the integral’s delayed failure.

How do Fourier transforms and the convolution theorem turn integrals into plateau/averaging behavior?

Fourier transforms relate the sinc function to the rect function: the engineer sinc (with

π

inside) transforms into a top-hat rect shape, and vice versa. A key identity links integrals over

(- \infty, \infty)

to evaluating the Fourier-transformed function at 0. Meanwhile, the convolution theorem says that multiplying functions in one domain corresponds to convolving them in the other. Since convolution with a rect-shaped kernel acts like a moving average, multiplying more sinc factors becomes equivalent to repeatedly averaging a rect plateau with shrinking windows—exactly the mechanism behind the moving-average threshold.

Review Questions

In the moving-average model, what condition ensures the value at $x = 0$ stays exactly at the plateau height after each averaging step?
How does changing the effective plateau length (from 1 to 2) alter the index where the sequence first drops below the constant value?
What roles do Fourier transform duality and the convolution theorem play in translating a sinc-product integral into an averaging/plateau picture?

Key Points

1
The Borwein sinc-product integrals equal $π$ exactly for many steps, then drop slightly below $π$ only after a specific threshold index (15 in the basic sequence, 113 when $2 cos (x)$ is included).
2
The “it must be a numerical bug” intuition fails because the eventual deviation has an exact closed-form value involving $π$ with extremely large integer numerator and denominator.
3
A moving-average model with a rect (step) function predicts the same breakpoints: the value at the center stays constant until shrinking averaging windows erode the plateau.
4
The break occurs when the sum of the selected odd reciprocal window widths exceeds the plateau length (1 for the basic setup, 2 for the modified setup).
5
Fourier transforms connect sinc and rect functions, turning integrals of sinc products into evaluations tied to rect-shaped objects.
6
The convolution theorem provides the mechanism: multiplying sinc factors corresponds to convolving/averaging rect functions, making the plateau-erosion explanation mathematically precise.

Highlights

The integrals remain pinned to π despite repeatedly multiplying in additional stretched sinc factors; the first failure happens at a precise index rather than gradually.

The moving-average analogy turns an analytic mystery into a geometric one: a center value stays at 1 until the averaging window no longer fits inside the plateau.

The delayed breakdown from 15 to 113 is explained by an effective doubling of plateau length, so the “eaten width” takes longer to exceed the threshold.

Fourier transform duality plus the convolution theorem converts sinc-product behavior into repeated averaging, aligning the integral sequence with the rect-plateau model.

Topics

Borwein integrals
Sinc and Rect
Fourier Transform
Convolution Theorem
Moving Average Plateau

Mentioned

Jonathan Borwein
David Borwein